This is a problem from a mathematics competition: Does there exist an irrational number $x$ such that the series $$\sum_{n=1}^{\infty}\{n!x\}<+\infty$$
where $\{ \}$ means the fractional part of a real number.
Someone gives a solution: first we can prove that every irrational number $x\in (0,1)$ can be written as the form uniquely $$x=\sum_{n=1}^{\infty}\frac{a_n}{n!}$$ where $a_n\in[0,n-1]$ is a integer. Next we can prove that the series $\sum_{n=1}^{\infty}\{n!x\}$ converges iff the series $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converges, and therefore we can construct some irrational numbers satisfying our condition.
My question is that whether all the irrational numbers $x$ which make the series $\sum_{n=1}^{\infty}\{n!x\}$ converge are transcendental numbers? My idea is that in order to make the series $\sum_{n=1}^{\infty}\frac{a_n}{n}$ converge, a lot of $a_n$ must be $0$. For example, if we let $a_n=1$ when $n$ is a power of $2$, and in other cases it is $0$, then I can prove $x=\sum_{n=1}^{\infty}\frac{a_n}{n!}$ is a transcendental number by using Liouville's theorem. In general cases I tried but failed to estimate how many $a_n$ cannot be $0$ in order to use Liouville's theorem again, could anyone give some help?
